How is a saturated solution in water prepared from a given solute?

Jul 15, 2017

Unknown.........

Explanation:

Saturation describes an equilibrium condition .......

And this specifies that the solution contains the SAME amount of solute as would be in equilibrium with undissolved solute. We might expect that as we increase the temperature of the solvent, we should increase the amount of solute it can dissolve..........but we have no data.

The definition of saturation is a problem area in A-level and 1st year chemistry.......... it DOES NOT MEAN the solvent can hold all the solute that it can.

Can you define $\text{supersaturation}$?

Jul 15, 2017

You can add 23.8 g more solute at 30.0 °C.

Explanation:

At 20.0 °C, you can dissolve

$\text{Mass of solute" = 75.0 color(red)(cancel(color(black)("g water"))) × "37.3 g solute"/(100 color(red)(cancel(color(black)("g water")))) = "27.98 g solute}$

At 30.0 °C you can dissolve

$\text{Mass of solute" = 75.0 color(red)(cancel(color(black)("g water"))) × "69.0 g solute"/(100 color(red)(cancel(color(black)("g water")))) = "51.75 g solute}$

$\text{Difference = (51.75 - 27.98) g = 23.8 g}$

You can dissolve 23.8 g more solute at 30.0 °C.